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# Please solve this for me..

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Guest Dec 31, 2014

#3
+92674
+10

Here's (c)

u(x,t) = cos(x + ct )

ut (x, t)  = - sin(x  ct) (c) = - (c)sin (x + ct)

So

utt = -(c)cos(x + ct)(c) = -(c2) cos (x + ct)

And

ux (x, t) = -sin(x + ct)(1)= -sin(x + ct)

So

uxx = -cos( x + t)(1) = -cos( x + ct)

So

(c2) uxx = (c2)[-cos( x + ct) ] = -(c2) cos (x + ct)

And ......we have shown that.....

utt = (c2) uxx

CPhill  Jan 1, 2015
#1
+92674
+10

∫√(4 - x^2) dx

Let x = 2sinΘ       dx = 2cosΘ dΘ  .....    when x =0, Θ = 0  and when x = 2, Θ = pi/2 ... so we have

∫√(4 - 4sin^2 Θ) 2cosΘ dΘ =

2∫√[(4(1 - sin^2  Θ) ] cos Θ  dΘ =

2 ∫2 √[ 1 - sin^2  Θ) ] cos Θ  dΘ =

4 ∫ cos Θ * cos Θ  dΘ =

4 ∫ cos ^2 Θ  dΘ          and using  cos^2 Θ = (1/2)(1 + cos2 Θ)  , we have

2 ∫ [1 + cos2 Θ]  dΘ  =

2Θ   + sin 2Θ  ....and substituting in the limits of integration, we have

2 [ pi/2 - 0 ] + [sin 2 (pi/2) - sin 2(0)] =

pi + [ sin (pi) - sin(0)] =

[pi ] +  [0]  =

CPhill  Dec 31, 2014
#2
+92674
+10

Here's (b)

q =  [P1P2 + 2P1 ] / [P1P2 - 2P2 ].......we can use the quotient rule here.....

∂q/∂P1 =  [ [ (P2 + 2)(P1P2 - 2P2) ] - [ (P1P2 + 2P1) (P2) ] ] /  [P1P2 - 2P2 ]2   =

[ P1P22 + 2P1P2 - 2P22 - 4P2  - P1P22 - 2P1P2 ] / [P1P2 - 2P2 ]2 =

- [ 2P22 + 4P2 ] /  [P1P2 - 2P2 ]2 =

-2P2 [ P2 + 2 ] / [P1P2 - 2P2 ]2

-2P2 [ P2 + 2 ] / [P2 (P1 - 2)]2 =

-2 [ P2 + 2 ] / [P2 (P1 - 2)2 ]

CPhill  Jan 1, 2015
#3
+92674
+10

Here's (c)

u(x,t) = cos(x + ct )

ut (x, t)  = - sin(x  ct) (c) = - (c)sin (x + ct)

So

utt = -(c)cos(x + ct)(c) = -(c2) cos (x + ct)

And

ux (x, t) = -sin(x + ct)(1)= -sin(x + ct)

So

uxx = -cos( x + t)(1) = -cos( x + ct)

So

(c2) uxx = (c2)[-cos( x + ct) ] = -(c2) cos (x + ct)

And ......we have shown that.....

utt = (c2) uxx

CPhill  Jan 1, 2015
#4
+808
+5

I know why your profile picture reads "666", because you're a devil with numbers ;)

Tetration  Jan 1, 2015
#5
+92674
0

LOL!!!! (I'm really not that good....more lucky....!!! )

CPhill  Jan 1, 2015
#6
+94105
+5

Thank you Chris

Please anon - put one question per post and turn your picture around the right way before you upload it.

If you are capable of calculus then you are able to learn how to rotate a picture!

Melody  Jan 2, 2015